Find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.) f(x) = x2 + 3x − 1

The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
  (-3/2 +3)(-3/2) -1 = -13/4

The one relative extreme is a minimum at (-1.5, -3.25).

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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
  x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.